An ionic compound $AB$ has a fluorite-type structure. If the radius of $B$ is $200$ pm, what is the ideal radius of $A$?


  • A: $82.8$ pm
  • B: $146.4$ pm
  • C: $40$ pm
  • D: $45$ pm

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The correct answer is: D

The solution can be understood as follows:

In an ideal fluorite structure, the cation ($A$) is present in a tetrahedral void. Hence, we can use the relationship for the ratio of ionic radii in this structure:

$$ \frac{r_c}{r_a} = 0.225 $$

Given that the radius of $B$ ($r_c$) is $200$ pm, we can solve for the radius of $A$ ($r_a$):

$$ r_a = \frac{r_c}{0.225} = \frac{200 \text{ pm}}{0.225} \approx 44.4 \text{ pm} $$

Since the closest option to this calculation is 45 pm, the ideal radius of $A$ is:

Option D: 45 pm

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